7.2
Eigenvectors of L2
so now we know that Lz Φm = mh̄Φm where Φm = (2π)−1/2 eimφ . But we also
know that there has to be a common set of eigenfunctions which are BOTH
eigenfunctions of Lz AND of L2 . We will call these Ylm (θ, φ). We already
know that these have to be eigenfunctions of Lz so
Lz Ylm (θφ) = mh̄Y (θφ)
but these must also be eigenfunctions of L2 so
L2 Ylm (θφ) = l(l + 1)h̄2 Ylm (θφ)
where again we have choosen to scale by h̄2 and the reason for calling the
eigenvalue l(l + 1) will become clear soon!
We can see that Ylm (θφ) must be separable into Θlm (θ)Φm (φ) where Φm is
as above and Θ can only be a function of θ and not φ as otherwise it would
∂
and then this wouldn’t be an eigenfunction of
be changed by Lz = −ih̄ ∂φ
both of them.
so solving
L2 Ylm (θφ) = l(l + 1)h̄2 Ylm (θφ)
L2 Θ(θ)Φm (φ) = l(l + 1)h̄2 Θ(θ)Φm (φ)
−h̄2
∂
1 ∂2 1 ∂
Θ(θ)Φm (φ) = l(l + 1)h̄2 Θ(θ)Φm (φ)
(sin θ ) +
sin θ ∂θ
∂θ
sin2 θ ∂φ2
Φ ∂
∂Θ
Θ ∂2Φ
(sin θ
)+
= −l(l + 1)Θ(θ)Φm (φ)
sin θ ∂θ
∂θ
sin2 θ ∂φ2
eigenfunctions of Lz had
∂Φm
= imΦm
∂φ
∂Φm
∂ 2 Φm
= im
= −m2 Φm
2
∂φ
∂φ
substitute in
Φ ∂
∂Θ
Θ
(sin θ
)−
m2 Φ = −l(l + 1)Θ(θ)Φm (φ)
sin θ ∂θ
∂θ
sin2 θ
1
divide out Φ to get just a function of θ so ∂/∂θ → d/dθ
dΘ
Θ
1 d
2
(sin θ
)−
2 m = −l(l + 1)Θ(θ)
sin θ dθ
dθ
sin θ
and rearrange
d
m2 1 d
(sin θ ) + l(l + 1) −
Θlm = 0
sin θ dθ
dθ
sin2 θ
the equation is ugly but well known mathematically. When m = 0 the solutions are Legendre polynomials Pl (cos θ), where the higest order polynomial
term is of order l where l = 0, 1, 2 . . . and is called the orbital angular momentum quantum number some examples are
P0 (cos θ) = 1
P1 (cos θ) = cos θ
1
P2 = (3 cos2 θ − 1)
2
1
P3 = (5 cos3 θ − 3 cos θ)
2
so like the hermite polynomials, the pattern is that even l only has even
powers of cos θ, odd l only has odd powers of cos θ.
for m 6= 0 the solutions are given by the associated Legendre polynomials
which are related to the |m|th derivative of Pl . But since Pl is a polynomial
of degree l, then its l+1 derivative will vanish. so for a fixed value of l, |m|
can only take the values 0 . . . l, so the allowed values of m for a given l are
m = −l, −l + 1, −l + 2 . . . 0, 1, 2 . . . (l − 1), l
so there are 2l+1 values of m for every l.
Then we normalise these Rto get our solutions for Θlm (θ). the normalisation
integral is only over θ so Θ∗lm (θ)Θlm (θ) sin θdθ. then we get
Θlm = (−1)m
2l + 1 (l
2
− m)! 1/2 m
Pl (cos θ) m ≥ 0
(l + m)!
Θlm = (−1)m Θl|m|
2
m<0
7.3
Spherical harmonics Ylm(θφ)
the eigenvectors common to L2 and Lz are given by
Ylm = Θlm Φm = (−1)m
2l + 1 (l − m)! 1/2
4π (l + m)!
∗
Ylm = (−1)m Yl,−m
Plm (cos θ)eimφ
m≥0
m<0
with allowed values m = 0, ±1, ±2 . . . ± l.
we can now see the limit on values which m can take more physically as
< L2 >=< L2x + L2y + L2z >=< L2x > + < L2y > + < L2z >
since these operators are all hermitian then these are all real so
< L2 > ≥ < L2z >
Z Z
Z Z
∗ 2
Ylm
L Ylm sin θdθdφ
≥
Z Z
∗
Ylm
l(l + 1)h̄2 Ylm sin θdθdφ ≥
2
l(l + 1)h̄ ≥
Z Z
∗ 2
Ylm
Lz Ylm sin θdθdφ
Z Z
∗
Ylm
Lz (mh̄Ylm ) sin θdθdφ
∗
Ylm
(m2 h̄2 Ylm ) sin θdθdφ
l(l + 1) ≥ m2
its obviously true for m = l, and not true for m = l + 1 as the rhs is l2 + 2l + 1
which is bigger than the lhs of l2 + l. so this limits the values of |m| ≤ l
The fact that we have determined Lz means that Lx and Ly cannot be simultaneously determined - if we measure them we will get a value which is
quantised at 0, ±h̄, ±2h̄ etc but the actual value is uncertain. However, we
can explicitally evaluate their averages, < Lx > and < Ly >. And these both
turn out to be zero. so although the particular value of Lx and Ly cannot be
predicted, their average value can.
So how do we visualise a system where we have L2 and Lz with quantised
values, but where < Lx > and < Ly >= 0? we can do this with a classical
3
q
vector model. The angular momentum vector of magnitude l(l + 1)h̄ precesses around the z axis, so the z component is always fixed at mh̄. Because
of the precession, < Lx > and < Ly > vanish.
so thats the agnular momentum done. what about the wavefunctions themselves? we can plots these Ylm out - A polar plot represents the magnitude
(absolute value) of the function as the length of a line centered at the origin,
and it represents the values of the angles θ, φ by the direction of the line.
but more useful than wavefunctions of course is probability distributions.
The probability of finding the particle within volume dV = dA = sin θdθdφ of
position θ, φ is dP = |Ylm |2 sin θdθdφ so if we want the probability density in φ
Rπ
then we have to integrate over all theta dP = D(φ)dφ = θ=0
|Ylm|2 sin θdθdφ
conversely,
if we wanted the probability density in θ we’d do dP = D(θ)dθ =
R 2π
2
|Y
|
dφ
sin θdφ similarly, if we want the probabilty per unit area on a
lm
0
sphere, then its dP dA = |Ylm |2 dA so
∗
D(A) = Ylm
Ylm = Θ∗lm Φ∗m Θlm Φm = (2π)−1 Θ∗lm Θlm
The φ dependence drops out, so the probability of finding the particle is
independent of φ so we can plot these more easily. This gives a polar diagram,
showing the dependence of the probability on θ. A polar plot represents the
magnitude (absolute value) of the function as the length of a line centered
at the origin, and it represents the values of the angle θ by the direction of
the line.
4